find the equation of the line passing through the point of intersection of 4x-5y =6 and 4x+3y+5=0 and perpendicular to the line x-3y+5=0

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Solution

$The given equations are : 4 x - 5 y = 6 . . . . . (1) 4 x + 3 y = - 5 . . . . . (2) Subtracting (2) from (1), we get - 8 y = 11 \Rightarrow y = - \frac{11}{8} Put y = - \frac{11}{8} in (1), we get 4 x - 5 \times (- \frac{11}{8}) = 6 \Rightarrow 4 x = 6 - \frac{55}{8} \Rightarrow 4 x = \frac{48 - 55}{8} \Rightarrow 4 x = - \frac{7}{8} \Rightarrow x = - \frac{7}{32} So, point of intersection of (1) and (2) is (- \frac{7}{32}, - \frac{11}{8}) Hence the required line passes through (- \frac{7}{32}, - \frac{11}{8}) . Now, the required line is ⊥ to the line x - 3 y + 5 = 0 . Now, x - 3 y + 5 = 0 \Rightarrow 3 y = 5 + x \Rightarrow y = \frac{5 + x}{3} Slope of the above line is \frac{1}{3} Now, slope of the required line = - 3 We know that equation of a line passing through (x_{1}, y_{1}) and having slope m is y - y_{1} = m (x - x_{1}) Now, equation of the required line is y - (- \frac{11}{8}) = - 3 [x - (- \frac{7}{32})] \Rightarrow y + \frac{11}{8} = - 3 (x + \frac{7}{32}) \Rightarrow \frac{8 y + 11}{8} = - 3 [\frac{32 x + 7}{32}] \Rightarrow 4 (8 y + 11) = - 3 (32 x + 7) \Rightarrow 32 y + 44 = - 96 x - 21 \Rightarrow 96 x + 32 y + 65 = 0$

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